Transformations and Distribution Function

#1
Hey guys,

Consider the following short example of transformations.

Let the joint density of \(x\ \text{and}\ y\) be given by the unit square, i.e.

\(\displaystyle f_{X,Y}\left( x,y \right)=\begin{cases} 1\ 0<x<1\ \text{and}\ 0<y<1 \\ 0\ \text{elsewhere}\end{cases} \)

Then the Cumulative Distribution Function of \(z=x+y\) is given by:

\(\displaystyle F_{Z}\left( z \right) = \begin{cases}0\ \text{for}\ z<0 \\ \int_0^z \int_0^{z-x} dydx\ \text{for}\ 0\leq z < 1 \\ 1-\int_{z-1}^1 \int_{z-x}^1 dydx\ \text{for}\ 1\leq z<2 \\ 1\ \text{for}\ 2\leq z \end{cases}\\)

I understand of course why the CDF is 0 and 1 but for the 2 cases in the middle I have been struggling to understand why we partition the CDF like that. In general what is the intuition behind the above result? I am very confused so any help is greatly appreciated, thanks.
 

BGM

TS Contributor
#2
Note that the CDF is \( F_Z(z) = \Pr\{X + Y \leq z\} \)

As what you said, the support of \( (X, Y) \) is a unit square in the \( x-y \) plane.

And the region \( \{(x, y): x + y \leq z\} \) is the bottom-left region divided by the line \( L: x + y = z \) which has a slope of \( -1 \).

So you just need to know the position of \( L \) for different values of \( z \). It will be easier if you can draw it out (at least in your mind)

E.g. when \( z < 0 \) or \( z > 2 \), \( L \) will not intersect with the unit square and thus you have the trivial solution.

And when \( z = 1 \), the line is just the diagonal line. Then you should understand why you need to split case at this value.

For the upper and lower limit it is just a result from "combining" the inequality, and you should be figure it out when you have the diagram.
 
#3
Thanks but I have to say that the second case, that is \(1-\int_{z-1}^1 \int_{z-x}^1 dydx\), is still far from evident to me. Could you please be a little more specific on that one?

Also, isn't that partition arbitrary? The function is continuous so what if we chose \(z=0.5\)?
 
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